control systems analysis and design by the root-locus method · 2018. 6. 20. · 270 chapter 6 /...

6

269

Control Systems Analysisand Design by the

Root-Locus Method

6–1 INTRODUCTION

The basic characteristic of the transient response of a closed-loop system is closelyrelated to the location of the closed-loop poles. If the system has a variable loop gain,then the location of the closed-loop poles depends on the value of the loop gain chosen.It is important, therefore, that the designer know how the closed-loop poles move inthe s plane as the loop gain is varied.

From the design viewpoint, in some systems simple gain adjustment may move theclosed-loop poles to desired locations.Then the design problem may become the selec-tion of an appropriate gain value. If the gain adjustment alone does not yield a desiredresult, addition of a compensator to the system will become necessary. (This subject isdiscussed in detail in Sections 6–6 through 6–9.)

The closed-loop poles are the roots of the characteristic equation. Finding the rootsof the characteristic equation of degree higher than 3 is laborious and will need computersolution. (MATLAB provides a simple solution to this problem.) However, just findingthe roots of the characteristic equation may be of limited value, because as the gain ofthe open-loop transfer function varies, the characteristic equation changes and thecomputations must be repeated.

A simple method for finding the roots of the characteristic equation has beendeveloped by W. R. Evans and used extensively in control engineering. This method,called the root-locus method, is one in which the roots of the characteristic equation

270 Chapter 6 / Control Systems Analysis and Design by the Root-Locus Method

H(s)

G(s)C(s)R(s)

+–

Figure 6–1Control system.

are plotted for all values of a system parameter. The roots corresponding to a par-ticular value of this parameter can then be located on the resulting graph. Note thatthe parameter is usually the gain, but any other variable of the open-loop transferfunction may be used. Unless otherwise stated, we shall assume that the gain of theopen-loop transfer function is the parameter to be varied through all values, from zeroto infinity.

By using the root-locus method the designer can predict the effects on the locationof the closed-loop poles of varying the gain value or adding open-loop poles and/oropen-loop zeros.Therefore, it is desired that the designer have a good understanding ofthe method for generating the root loci of the closed-loop system, both by hand and byuse of a computer software program like MATLAB.

In designing a linear control system, we find that the root-locus method proves to bequite useful, since it indicates the manner in which the open-loop poles and zeros shouldbe modified so that the response meets system performance specifications.This methodis particularly suited to obtaining approximate results very quickly.

Because generating the root loci by use of MATLAB is very simple, one may thinksketching the root loci by hand is a waste of time and effort. However, experience insketching the root loci by hand is invaluable for interpreting computer-generated rootloci, as well as for getting a rough idea of the root loci very quickly.

Outline of the Chapter. The outline of the chapter is as follows: Section 6–1 haspresented an introduction to the root-locus method. Section 6–2 details the conceptsunderlying the root-locus method and presents the general procedure for sketching rootloci using illustrative examples. Section 6–3 discusses generating root-locus plots withMATLAB. Section 6–4 treats a special case when the closed-loop system has positivefeedback. Section 6–5 presents general aspects of the root-locus approach to the designof closed-loop systems. Section 6–6 discusses the control systems design by lead com-pensation. Section 6–7 treats the lag compensation technique. Section 6–8 deals withthe lag–lead compensation technique. Finally, Section 6–9 discusses the parallel com-pensation technique.

6–2 ROOT-LOCUS PLOTS

Angle and Magnitude Conditions. Consider the negative feedback system shownin Figure 6–1. The closed-loop transfer function is

(6–1)C(s)

R(s)=

G(s)

1 + G(s)H(s)

Section 6–2 / Root-Locus Plots 271

The characteristic equation for this closed-loop system is obtained by setting thedenominator of the right-hand side of Equation (6–1) equal to zero. That is,

or

(6–2)

Here we assume that G(s)H(s) is a ratio of polynomials in s. [It is noted that wecan extend the analysis to the case when G(s)H(s) involves the transport lag e–Ts.]Since G(s)H(s) is a complex quantity, Equation (6–2) can be split into two equationsby equating the angles and magnitudes of both sides, respectively, to obtain thefollowing:

Angle condition:

(6–3)

Magnitude condition:

(6–4)

The values of s that fulfill both the angle and magnitude conditions are the roots ofthe characteristic equation, or the closed-loop poles. A locus of the points in thecomplex plane satisfying the angle condition alone is the root locus. The roots ofthe characteristic equation (the closed-loop poles) corresponding to a given valueof the gain can be determined from the magnitude condition. The details of applyingthe angle and magnitude conditions to obtain the closed-loop poles are presentedlater in this section.

In many cases, G(s)H(s) involves a gain parameter K, and the characteristic equa-tion may be written as

Then the root loci for the system are the loci of the closed-loop poles as the gain K isvaried from zero to infinity.

Note that to begin sketching the root loci of a system by the root-locus method wemust know the location of the poles and zeros of G(s)H(s). Remember that the anglesof the complex quantities originating from the open-loop poles and open-loop zeros tothe test point s are measured in the counterclockwise direction. For example, if G(s)H(s)is given by

G(s)H(s) =KAs + z1BAs + p1B As + p2B As + p3B As + p4B

1 +KAs + z1B As + z2B p As + zmBAs + p1B As + p2B p As + pnB = 0

∑G(s)H(s)∑ = 1

/G(s)H(s) = ;180°(2k + 1) (k = 0, 1, 2, p )

G(s)H(s) = -1

1 + G(s)H(s) = 0


Test point

Test point

–p4

–p3

–p2

–p1

s

s

–z1

f1

f1

jv

u4

u2

u3

u4u1

u3

u1

u2

�0–p4

–p2

A4

B1

A3

A2

A1

–p1

–p3

–z1

jv

�0

(b)(a)

where –p2 and –p3 are complex-conjugate poles, then the angle of G(s)H(s) is

where f1, u1, u2, u3, and u4 are measured counterclockwise as shown in Figures 6–2(a)and (b). The magnitude of G(s)H(s) for this system is

where A1 , A2 , A3 , A4 , and B1 are the magnitudes of the complex quantities s+p1,s+p2, s+p3, s+p4, and s+z1, respectively, as shown in Figure 6–2(a).

Note that, because the open-loop complex-conjugate poles and complex-conjugatezeros, if any, are always located symmetrically about the real axis, the root loci are alwayssymmetrical with respect to this axis.Therefore, we only need to construct the upper halfof the root loci and draw the mirror image of the upper half in the lower-half s plane.

Illustrative Examples. In what follows, two illustrative examples for constructingroot-locus plots will be presented. Although computer approaches to the constructionof the root loci are easily available, here we shall use graphical computation, combinedwith inspection, to determine the root loci upon which the roots of the characteristicequation of the closed-loop system must lie. Such a graphical approach will enhanceunderstanding of how the closed-loop poles move in the complex plane as the open-loop poles and zeros are moved.Although we employ only simple systems for illustrativepurposes, the procedure for finding the root loci is no more complicated for higher-order systems.

Because graphical measurements of angles and magnitudes are involved in the analy-sis, we find it necessary to use the same divisions on the abscissa as on the ordinate axiswhen sketching the root locus on graph paper.

∑G(s)H(s)∑ =KB1

A1 A2 A3 A4

/G(s)H(s) = f1 - u1 - u2 - u3 - u4

Figure 6–2(a) and (b) Diagramsshowing anglemeasurements fromopen-loop poles andopen-loop zero totest point s.


R(s) C(s)Ks(s + 1) (s + 2)

+–


EXAMPLE 6–1 Consider the negative feedback system shown in Figure 6–3. (We assume that the value of gainK is nonnegative.) For this system,

Let us sketch the root-locus plot and then determine the value of K such that the damping ratioz of a pair of dominant complex-conjugate closed-loop poles is 0.5.

For the given system, the angle condition becomes

The magnitude condition is

A typical procedure for sketching the root-locus plot is as follows:

1. Determine the root loci on the real axis. The first step in constructing a root-locus plot is tolocate the open-loop poles, s=0, s=–1, and s=–2, in the complex plane. (There are no open-loop zeros in this system.) The locations of the open-loop poles are indicated by crosses. (The lo-cations of the open-loop zeros in this book will be indicated by small circles.) Note that the startingpoints of the root loci (the points corresponding to K=0) are open-loop poles. The number ofindividual root loci for this system is three, which is the same as the number of open-loop poles.

To determine the root loci on the real axis, we select a test point, s. If the test point is on thepositive real axis, then

This shows that the angle condition cannot be satisfied. Hence, there is no root locus on the positivereal axis. Next, select a test point on the negative real axis between 0 and –1. Then

Thus

and the angle condition is satisfied.Therefore, the portion of the negative real axis between 0 and–1 forms a portion of the root locus. If a test point is selected between –1 and –2, then

and- /s - /s + 1 - /s + 2 = -360°

/s = /s + 1 = 180°, /s + 2 = 0°

- /s - /s + 1 - /s + 2 = -180°

/s = 180°, /s + 1 = /s + 2 = 0°

/s = /s + 1 = /s + 2 = 0°

∑G(s)∑ = 2 K

s(s + 1)(s + 2)2 = 1

= ;180°(2k + 1) (k = 0, 1, 2, p )

= - /s - /s + 1 - /s + 2

/G(s) = n K

s(s + 1)(s + 2)

G(s) =K

s(s + 1)(s + 2), H(s) = 1


It can be seen that the angle condition is not satisfied. Therefore, the negative real axis from –1to –2 is not a part of the root locus. Similarly, if a test point is located on the negative real axis from–2 to –q, the angle condition is satisfied. Thus, root loci exist on the negative real axis between0 and –1 and between –2 and –q.

2. Determine the asymptotes of the root loci. The asymptotes of the root loci as s approachesinfinity can be determined as follows: If a test point s is selected very far from the origin, then

and the angle condition becomes

or

Since the angle repeats itself as k is varied, the distinct angles for the asymptotes are determinedas 60°, –60°, and 180°. Thus, there are three asymptotes. The one having the angle of 180° is thenegative real axis.

Before we can draw these asymptotes in the complex plane, we must find the point wherethey intersect the real axis. Since

if a test point is located very far from the origin, then G(s) may be written as

For large values of s, this last equation may be approximated by

(6–5)

A root-locus diagram of G(s) given by Equation (6–5) consists of three straight lines.This can beseen as follows: The equation of the root locus is

or

which can be written as

/s + 1 = ;60°(2k + 1)

-3/s + 1 = ;180°(2k + 1)

n K(s + 1)3 = ;180°(2k + 1)

G(s) �K

(s + 1)3

G(s) =K

s3 + 3s2 + p

G(s) =K

s(s + 1)(s + 2)

Angles of asymptotes =;180°(2k + 1)

3 (k = 0, 1, 2, p )

-3/s = ;180°(2k + 1) (k = 0, 1, 2, p )

lims S q

G(s) = lims S q

K

s(s + 1)(s + 2)= lim

s S q

K

s3


jv

s

v = 0

–1

– j 3

j 3

s + 1 – 3v = 0

s + 1 + 3v = 0

0

Figure 6–4Three asymptotes.

By substituting s=s+jv into this last equation, we obtain

or

Taking the tangent of both sides of this last equation,


These three equations represent three straight lines, as shown in Figure 6–4.The three straight linesshown are the asymptotes. They meet at point s=–1. Thus, the abscissa of the intersection ofthe asymptotes and the real axis is obtained by setting the denominator of the right-hand sideof Equation (6–5) equal to zero and solving for s. The asymptotes are almost parts of the root lociin regions very far from the origin.

3. Determine the breakaway point. To plot root loci accurately, we must find the breakawaypoint, where the root-locus branches originating from the poles at 0 and –1 break away (as K isincreased) from the real axis and move into the complex plane.The breakaway point correspondsto a point in the s plane where multiple roots of the characteristic equation occur.

A simple method for finding the breakaway point is available. We shall present this methodin the following: Let us write the characteristic equation as

(6–6)f(s) = B(s) + KA(s) = 0

s + 1 -v

13= 0, s + 1 +

v

13= 0, v = 0

v

s + 1= 13 , -13 , 0

tan-1 v

s + 1= 60°, -60°, 0°

/s + jv + 1 = ;60°(2k + 1)


where A(s) and B(s) do not contain K. Note that f(s)=0 has multiple roots at points where

This can be seen as follows: Suppose that f(s) has multiple roots of order r, where .Then f(s)may be written as

Now we differentiate this equation with respect to s and evaluate df(s)/ds at s=s1. Then we get

(6–7)

This means that multiple roots of f(s) will satisfy Equation (6–7). From Equation (6–6), weobtain

(6–8)

where

The particular value of K that will yield multiple roots of the characteristic equation is obtainedfrom Equation (6–8) as

If we substitute this value of K into Equation (6–6), we get

or(6–9)

If Equation (6–9) is solved for s, the points where multiple roots occur can be obtained. On theother hand, from Equation (6–6) we obtain

and

If dK/ds is set equal to zero, we get the same equation as Equation (6–9). Therefore, the break-away points can be simply determined from the roots of

It should be noted that not all the solutions of Equation (6–9) or of dK/ds=0 correspond toactual breakaway points. If a point at which dK/ds=0 is on a root locus, it is an actual breakawayor break-in point. Stated differently, if at a point at which dK/ds=0 the value of K takes a realpositive value, then that point is an actual breakaway or break-in point.

dKds

= 0

dKds

= -B¿(s)A(s) - B(s)A¿(s)

A2(s)

K = -B(s)

A(s)

B(s)A¿(s) - B¿(s)A(s) = 0

f(s) = B(s) -B¿(s)

A¿(s)A(s) = 0

K = -B¿(s)

A¿(s)

A¿(s) =dA(s)

ds, B¿(s) =

dB(s)

ds

df(s)

ds= B¿(s) + KA¿(s) = 0

df(s)

ds2s = s1

= 0

f(s) = As - s1BrAs - s2B p As - snBr � 2

df(s)

ds= 0


For the present example, the characteristic equation G(s)+1=0 is given by

or

By setting dK/ds=0, we obtain

or

Since the breakaway point must lie on a root locus between 0 and –1, it is clear that s=–0.4226corresponds to the actual breakaway point. Point s=–1.5774 is not on the root locus. Hence, thispoint is not an actual breakaway or break-in point. In fact, evaluation of the values of K corre-sponding to s=–0.4226 and s=–1.5774 yields

4. Determine the points where the root loci cross the imaginary axis. These points can be foundby use of Routh’s stability criterion as follows: Since the characteristic equation for the presentsystem is

the Routh array becomes

The value of K that makes the s1 term in the first column equal zero is K=6. The crossing pointson the imaginary axis can then be found by solving the auxiliary equation obtained from the s2

row; that is,

which yields

The frequencies at the crossing points on the imaginary axis are thus The gain valuecorresponding to the crossing points is K=6.

An alternative approach is to let s=jv in the characteristic equation, equate both the realpart and the imaginary part to zero, and then solve for v and K. For the present system, the char-acteristic equation, with s=jv, is

or

Equating both the real and imaginary parts of this last equation to zero, respectively, we obtain

K - 3v2 = 0, 2v - v3 = 0

AK - 3v2B + jA2v - v3B = 0

(jv)3 + 3(jv)2 + 2(jv) + K = 0

v = ;12 .

s = ;j12

3s2 + K = 3s2 + 6 = 0

s3

s2

s1

s0

13

6 - K

3K

2K

s3 + 3s2 + 2s + K = 0

K = -0.3849, for s = -1.5774

K = 0.3849, for s = -0.4226

s = -0.4226, s = -1.5774

dKds

= - A3s2 + 6s + 2B = 0

K = - As3 + 3s2 + 2sBK

s(s + 1)(s + 2)+ 1 = 0


jv

j1

– j1

–1–2

s + 1s + 2

u2

u1

u3

0

s

s

Figure 6–5Construction of rootlocus.

jv

j1

– j1

1–2–3 0 s

K = 6

K = 6

K = 1.0383

K = 1.0383

K`

K`

j2

– j2

60°

–1

Figure 6–6Root-locus plot.

from which

Thus, root loci cross the imaginary axis at and the value of K at the crossing points is 6.Also, a root-locus branch on the real axis touches the imaginary axis at v=0. The value of K iszero at this point.

5. Choose a test point in the broad neighborhood of the jv axis and the origin, as shown inFigure 6–5, and apply the angle condition. If a test point is on the root loci, then the sum of thethree angles, u1+u2+u3 , must be 180°. If the test point does not satisfy the angle condition,select another test point until it satisfies the condition. (The sum of the angles at the test point willindicate the direction in which the test point should be moved.) Continue this process and locatea sufficient number of points satisfying the angle condition.

6. Draw the root loci, based on the information obtained in the foregoing steps, as shown inFigure 6–6.

v = ;12 ,

v = ;12 , K = 6 or v = 0, K = 0


7. Determine a pair of dominant complex-conjugate closed-loop poles such that the dampingratio z is 0.5. Closed-loop poles with z=0.5 lie on lines passing through the origin and makingthe angles with the negative real axis. From Figure 6–6, such closed-loop poles having z=0.5 are obtained as follows:

The value of K that yields such poles is found from the magnitude condition as follows:

Using this value of K, the third pole is found at s=–2.3326.Note that, from step 4, it can be seen that for K=6 the dominant closed-loop poles lie on the

imaginary axis at With this value of K, the system will exhibit sustained oscillations.For K>6, the dominant closed-loop poles lie in the right-half s plane, resulting in an unstablesystem.

Finally, note that, if necessary, the root loci can be easily graduated in terms of K by use of themagnitude condition. We simply pick out a point on a root locus, measure the magnitudes of thethree complex quantities s, s+1, and s+2, and multiply these magnitudes; the product is equalto the gain value K at that point, or

Graduation of the root loci can be done easily by use of MATLAB. (See Section 6–3.)

EXAMPLE 6–2 In this example, we shall sketch the root-locus plot of a system with complex-conjugate open-loop poles. Consider the negative feedback system shown in Figure 6–7. For this system,

where K � 0. It is seen that G(s) has a pair of complex-conjugate poles at

A typical procedure for sketching the root-locus plot is as follows:

1. Determine the root loci on the real axis. For any test point s on the real axis, the sum of theangular contributions of the complex-conjugate poles is 360°, as shown in Figure 6–8.Thus the neteffect of the complex-conjugate poles is zero on the real axis.The location of the root locus on thereal axis is determined from the open-loop zero on the negative real axis.A simple test reveals thata section of the negative real axis, that between –2 and –q, is a part of the root locus. It is notedthat, since this locus lies between two zeros (at s=–2 and s=–q), it is actually a part of tworoot loci, each of which starts from one of the two complex-conjugate poles. In other words, tworoot loci break in the part of the negative real axis between –2 and –q.

s = -1 + j12 , s = -1 - j12

G(s) =K(s + 2)

s2 + 2s + 3, H(s) = 1

∑s∑ � ∑s + 1∑ � ∑s + 2∑ = K

s = ;j12 .

= 1.0383

K = ∑s(s + 1)(s + 2)∑s = -0.3337 + j0.5780

s1 = -0.3337 + j0.5780, s2 = -0.3337 - j0.5780

;cos-1 z = ;cos-1 0.5 = ;60°

R(s) C(s)K(s + 2)s2 + 2s + 3

+–



jv

0 s

u92

s

–p2

f1

f91

–z1

–p1

u2

u1

Figure 6–9Determination of theangle of departure.

Since there are two open-loop poles and one zero, there is one asymptote, which coincides withthe negative real axis.

2. Determine the angle of departure from the complex-conjugate open-loop poles. The pres-ence of a pair of complex-conjugate open-loop poles requires the determination of the angle ofdeparture from these poles. Knowledge of this angle is important, since the root locus near a com-plex pole yields information as to whether the locus originating from the complex pole migratestoward the real axis or extends toward the asymptote.

Referring to Figure 6–9, if we choose a test point and move it in the very vicinity of the com-plex open-loop pole at s=–p1, we find that the sum of the angular contributions from the poleat s=p2 and zero at s=–z1 to the test point can be considered remaining the same. If the testpoint is to be on the root locus, then the sum of –u1 , and must be wherek=0, 1, 2, p . Thus, in the example,

or

The angle of departure is then

u1 = 180° - u2 + f1 = 180° - 90° + 55° = 145°

u1 = 180° - uœ2 + fœ

1 = 180° - u2 + f1

fœ1 - Au1 + uœ

2B = ;180°(2k + 1)

;180°(2k + 1),-uœ2fœ

1 ,

jv

–1 0–2 s

j 2

– j 2

Testpoint u2

u1

Figure 6–8Determination of theroot locus on the realaxis.


jv

j1

– j1

1–2–3–4 0 s

z = 0.7 line

j2

– j2

145°

–1

Figure 6–10Root-locus plot.

Since the root locus is symmetric about the real axis, the angle of departure from the pole ats=–p2 is –145°.

3. Determine the break-in point. A break-in point exists where a pair of root-locus branchescoalesces as K is increased. For this problem, the break-in point can be found as follows: Since

we have

which gives

or

Notice that point s=–3.7320 is on the root locus. Hence this point is an actual break-in point.(Note that at point s=–3.7320 the corresponding gain value is K=5.4641.) Since points=–0.2680 is not on the root locus, it cannot be a break-in point. (For point s=–0.2680, the cor-responding gain value is K=–1.4641.)

4. Sketch a root-locus plot, based on the information obtained in the foregoing steps. Todetermine accurate root loci, several points must be found by trial and error between the break-in point and the complex open-loop poles. (To facilitate sketching the root-locus plot, we shouldfind the direction in which the test point should be moved by mentally summing up the changeson the angles of the poles and zeros.) Figure 6–10 shows a complete root-locus plot for the systemconsidered.

s = -3.7320 or s = -0.2680

s2 + 4s + 1 = 0

dKds

= -(2s + 2)(s + 2) - As2 + 2s + 3B

(s + 2)2 = 0

K = -s2 + 2s + 3

s + 2


The value of the gain K at any point on root locus can be found by applying the magnitudecondition or by use of MATLAB (see Section 6–3). For example, the value of K at which thecomplex-conjugate closed-loop poles have the damping ratio z=0.7 can be found by locating theroots, as shown in Figure 6–10, and computing the value of K as follows:

Or use MATLAB to find the value of K. (See Section 6–4.)It is noted that in this system the root locus in the complex plane is a part of a circle. Such a

circular root locus will not occur in most systems. Circular root loci may occur in systems that in-volve two poles and one zero, two poles and two zeros, or one pole and two zeros. Even in suchsystems, whether circular root loci occur depends on the locations of poles and zeros involved.

To show the occurrence of a circular root locus in the present system, we need to derive theequation for the root locus. For the present system, the angle condition is

If s=s+jv is substituted into this last equation, we obtain


or

Taking tangents of both sides of this last equation using the relationship

(6–10)

we obtain

or

which can be simplified to

or

This last equation is equivalent to

or (s + 2)2 + v2 = A13B2v = 0

v C(s + 2)2 + v2 - 3 D = 0

2v(s + 1)

(s + 1)2 - Av2 - 2B =v

s + 2

v - 12

s + 1+v + 12

s + 1

1 - a v - 12

s + 1b a v + 12

s + 1b =

v

s + 2; 0

1 <v

s + 2* 0

tan c tan-1 a v - 12

s + 1b + tan-1 a v + 12

s + 1b d = tan c tan-1 a v

s + 2b ; 180°(2k + 1) d

tan (x ; y) =tan x ; tan y

1 < tan x tan y

tan-1 a v - 12

s + 1b + tan-1 a v + 12

s + 1b = tan-1 a v

s + 2b ; 180°(2k + 1)

tan-1 a v

s + 2b - tan-1 a v - 12

s + 1b - tan-1 a v + 12

s + 1b = ;180°(2k + 1)

/s + 2 + jv - /s + 1 + jv - j12 - /s + 1 + jv + j12 = ;180°(2k + 1)

/s + 2 - /s + 1 - j12 - /s + 1 + j12 = ;180°(2k + 1)

K = 2 As + 1 - j12B As + 1 + j12Bs + 2

2s = -1.67 + j1.70

= 1.34


These two equations are the equations for the root loci for the present system. Notice that the firstequation, v=0, is the equation for the real axis. The real axis from s=–2 to s=–q corre-sponds to a root locus for K � 0. The remaining part of the real axis corresponds to a root locuswhen K is negative. (In the present system, K is nonnegative.) (Note that K < 0 corresponds tothe positive-feedback case.) The second equation for the root locus is an equation of a circle withcenter at s=–2, v=0 and the radius equal to That part of the circle to the left of thecomplex-conjugate poles corresponds to a root locus for K � 0. The remaining part of the circlecorresponds to a root locus when K is negative.

It is important to note that easily interpretable equations for the root locus can be derived forsimple systems only. For complicated systems having many poles and zeros, any attempt to deriveequations for the root loci is discouraged. Such derived equations are very complicated and theirconfiguration in the complex plane is difficult to visualize.

General Rules for Constructing Root Loci. For a complicated system with manyopen-loop poles and zeros, constructing a root-locus plot may seem complicated, butactually it is not difficult if the rules for constructing the root loci are applied. By locat-ing particular points and asymptotes and by computing angles of departure from com-plex poles and angles of arrival at complex zeros, we can construct the general form ofthe root loci without difficulty.

We shall now summarize the general rules and procedure for constructing the rootloci of the negative feedback control system shown in Figure 6–11.

First, obtain the characteristic equation

Then rearrange this equation so that the parameter of interest appears as the multiply-ing factor in the form

(6–11)

In the present discussions, we assume that the parameter of interest is the gain K, whereK>0. (If K<0, which corresponds to the positive-feedback case, the angle condi-tion must be modified. See Section 6–4.) Note, however, that the method is still appli-cable to systems with parameters of interest other than gain. (See Section 6–6.)

1. Locate the poles and zeros of G(s)H(s) on the s plane.The root-locus branches startfrom open-loop poles and terminate at zeros (finite zeros or zeros at infinity). From thefactored form of the open-loop transfer function, locate the open-loop poles and zerosin the s plane. CNote that the open-loop zeros are the zeros of G(s)H(s), while theclosed-loop zeros consist of the zeros of G(s) and the poles of H(s). D

1 +KAs + z1B As + z2B p As + zmBAs + p1B As + p2B p As + pnB = 0

1 + G(s)H(s) = 0

13 .

H(s)

G(s)C(s)R(s)

+–



Note that the root loci are symmetrical about the real axis of the s plane, because thecomplex poles and complex zeros occur only in conjugate pairs.

A root-locus plot will have just as many branches as there are roots of the character-istic equation. Since the number of open-loop poles generally exceeds that of zeros, thenumber of branches equals that of poles. If the number of closed-loop poles is the sameas the number of open-loop poles, then the number of individual root-locus branchesterminating at finite open-loop zeros is equal to the number m of the open-loop zeros.The remaining n-m branches terminate at infinity (n-m implicit zeros at infinity)along asymptotes.

If we include poles and zeros at infinity, the number of open-loop poles is equalto that of open-loop zeros. Hence we can always state that the root loci start at thepoles of G(s)H(s) and end at the zeros of G(s)H(s), as K increases from zero to in-finity, where the poles and zeros include both those in the finite s plane and those atinfinity.

2. Determine the root loci on the real axis. Root loci on the real axis are determinedby open-loop poles and zeros lying on it. The complex-conjugate poles and complex-conjugate zeros of the open-loop transfer function have no effect on the location of theroot loci on the real axis because the angle contribution of a pair of complex-conjugatepoles or complex-conjugate zeros is 360° on the real axis. Each portion of the rootlocus on the real axis extends over a range from a pole or zero to another pole or zero.In constructing the root loci on the real axis, choose a test point on it. If the total num-ber of real poles and real zeros to the right of this test point is odd, then this point lieson a root locus. If the open-loop poles and open-loop zeros are simple poles and sim-ple zeros, then the root locus and its complement form alternate segments along thereal axis.

3. Determine the asymptotes of root loci. If the test point s is located far from the ori-gin, then the angle of each complex quantity may be considered the same. One open-loopzero and one open-loop pole then cancel the effects of the other. Therefore, the rootloci for very large values of s must be asymptotic to straight lines whose angles (slopes)are given by

where number of finite poles of G(s)H(s)

number of finite zeros of G(s)H(s)

Here, k=0 corresponds to the asymptotes with the smallest angle with the real axis.Al-though k assumes an infinite number of values, as k is increased the angle repeats itself,and the number of distinct asymptotes is n-m.

All the asymptotes intersect at a point on the real axis. The point at which they doso is obtained as follows: If both the numerator and denominator of the open-loop trans-fer function are expanded, the result is

G(s)H(s) =K Csm + Az1 + z2 + p + zmBsm - 1 + p + z1 z2

p zm Dsn + Ap1 + p2 + p + pnBsn - 1 + p + p1 p2

p pn

m =

n =

Angles of asymptotes =;180°(2k + 1)

n - m (k = 0, 1, 2, p )


If a test point is located very far from the origin, then by dividing the denominator bythe numerator, it is possible to write G(s)H(s) as

or

(6–12)

The abscissa of the intersection of the asymptotes and the real axis is then obtained bysetting the denominator of the right-hand side of Equation (6–12) equal to zero andsolving for s, or

(6–13)

[Example 6–1 shows why Equation (6–13) gives the intersection.] Once this intersectionis determined, the asymptotes can be readily drawn in the complex plane.

It is important to note that the asymptotes show the behavior of the root loci forA root-locus branch may lie on one side of the corresponding asymptote or may

cross the corresponding asymptote from one side to the other side.

4. Find the breakaway and break-in points. Because of the conjugate symmetry ofthe root loci, the breakaway points and break-in points either lie on the real axis oroccur in complex-conjugate pairs.

If a root locus lies between two adjacent open-loop poles on the real axis, then thereexists at least one breakaway point between the two poles. Similarly, if the root locus liesbetween two adjacent zeros (one zero may be located at –q) on the real axis, then therealways exists at least one break-in point between the two zeros. If the root locus lies be-tween an open-loop pole and a zero (finite or infinite) on the real axis, then there mayexist no breakaway or break-in points or there may exist both breakaway and break-inpoints.

Suppose that the characteristic equation is given by

The breakaway points and break-in points correspond to multiple roots of the charac-teristic equation. Hence, as discussed in Example 6–1, the breakaway and break-in pointscan be determined from the roots of

(6–14)

where the prime indicates differentiation with respect to s. It is important to note thatthe breakaway points and break-in points must be the roots of Equation (6–14), but notall roots of Equation (6–14) are breakaway or break-in points. If a real root of Equation(6–14) lies on the root-locus portion of the real axis, then it is an actual breakaway orbreak-in point. If a real root of Equation (6–14) is not on the root-locus portion of thereal axis, then this root corresponds to neither a breakaway point nor a break-in point.

dKds

= -B¿(s)A(s) - B(s)A¿(s)

A2(s)= 0

B(s) + KA(s) = 0

∑s∑ � 1.

s = -Ap1 + p2 + p + pnB - Az1 + z2 + p + zmB

n - m

G(s)H(s) =Kc s +

Ap1 + p2 + p + pnB - Az1 + z2 + p + zmBn - m

d n - m

G(s)H(s) =K

sn - m + C Ap1 + p2 + p + pnB - Az1 + z2 + p + zmB Dsn - m - 1 + p


jv

s

Angle ofdeparture

u2

u1f

0Figure 6–12Construction of theroot locus. [Angle ofdeparture=180°-(u1+u2 ) +f.]

If two roots s=s1 and s=–s1 of Equation (6–14) are a complex-conjugate pair and ifit is not certain whether they are on root loci, then it is necessary to check the corre-sponding K value. If the value of K corresponding to a root s=s1 of is pos-itive, point s=s1 is an actual breakaway or break-in point. (Since K is assumed to benonnegative, if the value of K thus obtained is negative, or a complex quantity, thenpoint s=s1 is neither a breakaway nor a break-in point.)

5. Determine the angle of departure (angle of arrival) of the root locus from a com-plex pole (at a complex zero). To sketch the root loci with reasonable accuracy, we mustfind the directions of the root loci near the complex poles and zeros. If a test point is cho-sen and moved in the very vicinity of a complex pole (or complex zero), the sum of theangular contributions from all other poles and zeros can be considered to remain thesame. Therefore, the angle of departure (or angle of arrival) of the root locus from acomplex pole (or at a complex zero) can be found by subtracting from 180° the sum ofall the angles of vectors from all other poles and zeros to the complex pole (or complexzero) in question, with appropriate signs included.

Angle of departure from a complex pole=180°– (sum of the angles of vectors to a complex pole in question from other poles)± (sum of the angles of vectors to a complex pole in question from zeros)

Angle of arrival at a complex zero=180°– (sum of the angles of vectors to a complex zero in question from other zeros)± (sum of the angles of vectors to a complex zero in question from poles)

The angle of departure is shown in Figure 6–12.

6. Find the points where the root loci may cross the imaginary axis. The points wherethe root loci intersect the jv axis can be found easily by (a) use of Routh’s stability cri-terion or (b) letting s=jv in the characteristic equation, equating both the real part andthe imaginary part to zero, and solving for v and K.The values of v thus found give thefrequencies at which root loci cross the imaginary axis. The K value corresponding toeach crossing frequency gives the gain at the crossing point.

7. Taking a series of test points in the broad neighborhood of the origin of the s plane,sketch the root loci. Determine the root loci in the broad neighborhood of the jv axisand the origin. The most important part of the root loci is on neither the real axis northe asymptotes but is in the broad neighborhood of the jv axis and the origin.The shape

dK�ds = 0


of the root loci in this important region in the s plane must be obtained with reasonableaccuracy. (If accurate shape of the root loci is needed, MATLAB may be used rather thanhand calculations of the exact shape of the root loci.)

8. Determine closed-loop poles. A particular point on each root-locus branch will bea closed-loop pole if the value of K at that point satisfies the magnitude condition. Con-versely, the magnitude condition enables us to determine the value of the gain K at anyspecific root location on the locus. (If necessary, the root loci may be graduated in termsof K. The root loci are continuous with K.)

The value of K corresponding to any point s on a root locus can be obtained usingthe magnitude condition, or

This value can be evaluated either graphically or analytically. (MATLAB can be usedfor graduating the root loci with K. See Section 6–3.)

If the gain K of the open-loop transfer function is given in the problem, then by ap-plying the magnitude condition, we can find the correct locations of the closed-looppoles for a given K on each branch of the root loci by a trial-and-error approach or byuse of MATLAB, which will be presented in Section 6–3.

Comments on the Root-Locus Plots. It is noted that the characteristic equa-tion of the negative feedback control system whose open-loop transfer function is

is an nth-degree algebraic equation in s. If the order of the numerator of G(s)H(s) islower than that of the denominator by two or more (which means that there are two ormore zeros at infinity), then the coefficient a1 is the negative sum of the roots of theequation and is independent of K. In such a case, if some of the roots move on the locustoward the left as K is increased, then the other roots must move toward the right as Kis increased. This information is helpful in finding the general shape of the root loci.

It is also noted that a slight change in the pole–zero configuration may cause signif-icant changes in the root-locus configurations. Figure 6–13 demonstrates the fact that aslight change in the location of a zero or pole will make the root-locus configurationlook quite different.

G(s)H(s) =KAsm + b1 sm - 1 + p + bmB

sn + a1 sn - 1 + p + an

(n � m)

K =product of lengths between point s to poles

product of lengths between point s to zeros

jv

s

jv

s

Figure 6–13Root-locus plots.


C(s)R(s)

(a)

1s

K(s + 1) (s + 2)

C(s)R(s)

(c)

1s + 1

Ks(s + 1) (s + 2)

Ks(s + 2)

s + 1

H(s)

C(s)R(s)

(b)

+–

+–

+–

G(s)

+–

Figure 6–14(a) Control systemwith velocityfeedback; (b) and (c) modified blockdiagrams.

Cancellation of Poles of G(s) with Zeros of H(s). It is important to note thatif the denominator of G(s) and the numerator of H(s) involve common factors, then thecorresponding open-loop poles and zeros will cancel each other, reducing the degree ofthe characteristic equation by one or more. For example, consider the system shown inFigure 6–14(a). (This system has velocity feedback.) By modifying the block diagram ofFigure 6–14(a) to that shown in Figure 6–14(b), it is clearly seen that G(s) and H(s)have a common factor s+1. The closed-loop transfer function C(s)/R(s) is

The characteristic equation is

Because of the cancellation of the terms (s+1) appearing in G(s) and H(s), however,we have

The reduced characteristic equation is

The root-locus plot of G(s)H(s) does not show all the roots of the characteristic equa-tion, only the roots of the reduced equation.

To obtain the complete set of closed-loop poles, we must add the canceled pole ofG(s)H(s) to those closed-loop poles obtained from the root-locus plot of G(s)H(s).The important thing to remember is that the canceled pole of G(s)H(s) is a closed-looppole of the system, as seen from Figure 6–14(c).

s(s + 2) + K = 0

=s(s + 2) + K

s(s + 2)

1 + G(s)H(s) = 1 +K(s + 1)

s(s + 1)(s + 2)

Cs(s + 2) + K D(s + 1) = 0

C(s)

R(s)=

K

s(s + 1)(s + 2) + K(s + 1)


Typical Pole–Zero Configurations and Corresponding Root Loci. In summa-rizing, we show several open-loop pole–zero configurations and their correspondingroot loci in Table 6–1. The pattern of the root loci depends only on the relative separa-tion of the open-loop poles and zeros. If the number of open-loop poles exceeds thenumber of finite zeros by three or more, there is a value of the gain K beyond which rootloci enter the right-half s plane, and thus the system can become unstable.A stable sys-tem must have all its closed-loop poles in the left-half s plane.

Table 6–1 Open-Loop Pole–Zero Configurations and the Corresponding Root Loci

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

jv

s

control systems analysis and design by the root-locus method · 2018. 6. 20. · 270 chapter 6 /...

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